In an arithmetic progression (A.P), $a_7 = 9$. At what value of its difference is the product $a_1 \cdot a_2 \cdot a_7$ the least?
Let the first term of A.P $= a_1$ and the common difference $= d$
$n^{th}$ term of A.P $= a_n = a_1 + \left(n - 1\right) d$
$\therefore \;$ $2^{nd}$ term of A.P $= a_2 = a_1 + d$ $\;\;\; \cdots \; (1)$
and $\;$ $7^{th}$ term of A.P $= a_7 = a_1 + 6d$
Given: $\;$ $a_7 = 9$ $\;\;\; \cdots \; (2)$
i.e. $\;$ $a_1 + 6d = 9$ $\implies$ $a_1 = 9 - 6d$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(1)$ becomes
$a_2 = 9 - 6d + d = 9 - 5d$ $\;\;\; \cdots \; (4)$
From equations $(2)$, $(3)$ and $(4)$, the product
$\begin{aligned}
P = a_1 \times a_2 \times a_7 & = \left(9 - 6d\right) \times \left(9 - 5d\right) \times 9 \\\\
& = 9 \times \left(81 - 99d + 30 d^2\right)
\end{aligned}$
For the product $P$ to be the least, $\;$ $\dfrac{d\left(P\right)}{d \left(d\right)} = 0$
i.e. $\;$ $\dfrac{d\left(P\right)}{d \left(d\right)} = 9 \times \left(-99 + 60d\right) = 0$
$\implies$ $d = \dfrac{99}{60} = \dfrac{33}{20}$
$\therefore \;$ When the common difference of the A.P is $d = \dfrac{33}{20}$, the product $a_1 \times a_2 \times a_7$ is the least.